The given function is $f(x) = x^2 + 6x + 3$. To complete parts a through c, I'll assume that typical questions in such problems involve:

1. Finding the vertex of the parabola.
2. Finding the axis of symmetry.
3. Identifying the intercepts (both the x-intercepts and the y-intercept).

a. Finding the vertex

The vertex form of a quadratic function is given by: $f(x) = a(x - h)^2 + k$ where $(h, k)$ is the vertex of the parabola.

For a quadratic function in standard form $f(x) = ax^2 + bx + c$, the x-coordinate of the vertex is given by: $x = -\frac{b}{2a}$ In this case, $a = 1$, $b = 6$, and $c = 3$. Substituting the values: $x = -\frac{6}{2(1)} = -3$

To find the y-coordinate of the vertex, substitute $x = -3$ into the original function: $f(-3) = (-3)^2 + 6(-3) + 3 = 9 - 18 + 3 = -6$ Thus, the vertex is at $(-3, -6)$.

b. Axis of Symmetry

The axis of symmetry is the vertical line that passes through the vertex. Its equation is: $x = -3$

c. Intercepts

1. Y-intercept: The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function: $f(0) = 0^2 + 6(0) + 3 = 3$ So, the y-intercept is $(0, 3)$.

2. X-intercepts: To find the x-intercepts, we set $f(x) = 0$: $x^2 + 6x + 3 = 0$ We can solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Substituting $a = 1$, $b = 6$, and $c = 3$: $x = \frac{-6 \pm \sqrt{6^2 - 4(1)(3)}}{2(1)} = \frac{-6 \pm \sqrt{36 - 12}}{2} = \frac{-6 \pm \sqrt{24}}{2} = \frac{-6 \pm 2\sqrt{6}}{2}$ $x = -3 \pm \sqrt{6}$ Thus, the x-intercepts are $x = -3 + \sqrt{6}$ and $x = -3 - \sqrt{6}$.

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Related Questions:

1. How do we convert a quadratic function into vertex form?
2. What does the discriminant tell us about the number of x-intercepts?
3. How can we graph a quadratic function using just its vertex and intercepts?
4. What is the significance of the axis of symmetry in the graph of a parabola?
5. How do transformations (shifting, reflecting) affect the graph of a quadratic function?

Tip:

For any quadratic function $ax^2 + bx + c$, the discriminant $\Delta = b^2 - 4ac$ helps determine the nature of the roots. If $\Delta > 0$, the quadratic has two distinct real roots, if $\Delta = 0$, there is one real root (the vertex), and if $\Delta < 0$, there are no real roots (the parabola doesn't intersect the x-axis).